Question

Find the area of the parallelogram with vertices: P(0,0,0), Q(-4,3,-1), R(-4,2,-2), S(-8,5,-3).

Answer 1

Explanation:

Co-ordinate of point P(0,0,0)

Q (-4,3,-1)

R (-4,2,-2)

S (-8,5,-3)

`\vec{PQ}=<-4,3,-1>`

`\vec{PS}=<-8,5,-3>`

Area of Parallelogram
`=|\vec{PQ}* \vec{PS}|`

`Area=\begin{vmatrix}i &j &k \\ -4 &3 &-1 \\ -8 &5 &-3 \end{vmatrix}`

`A=|\hat{i}(-9+5)-\hat{j}(-12-8)+\hat{k}(-20+24)|`

`A=|-4\hat{i}+20\hat{j}+4\hat{k}|`

`A=√((-4)^2+(20)^2+(4)^2)`

`A=√(432)`

`A=20.78\ unit^2`